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please explain thanks very much

 Apr 17, 2019
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\( \displaystyle\int_c^d\;\frac{1}{2x \sqrt x}\;dx\\ =\frac{1}{2} \displaystyle\int_c^d\;x ^{(\frac{-3}{2})}\;dx\\ =\frac{1}{2} \displaystyle \left[\;-2x ^{(\frac{-1}{2})}\right]_c^d\\ =-1 \displaystyle \left[\;x ^{(\frac{-1}{2})}\right]_c^d\\ =-1 \displaystyle \left[\;\frac{1}{\sqrt x}\right]_c^d\\ =-1 \displaystyle \left[\;\frac{1}{\sqrt d}-\frac{1}{\sqrt c}\right]\\ =\;\dfrac{1}{\sqrt c}-\dfrac{1}{\sqrt d} \)

 

\( \displaystyle\int_0^9\;\frac{1}{2x \sqrt x}\;dx\\ =\dfrac{1}{\sqrt 0}-\dfrac{1}{\sqrt 9}\\ =\;undefined\)

 

 

 

\( \displaystyle\int_9^\infty \;\frac{1}{2x \sqrt x}\;dx\\ =\dfrac{1}{\sqrt 9}-\dfrac{1}{\sqrt \infty}\\ =\dfrac{1}{3}\\ \)

 

I have not checked this so you need to do so. 

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 Apr 17, 2019
edited by Melody  Apr 17, 2019

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